LeetCode 50. Pow(x, n)x的n的次方

Implement pow(x, n).

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public class Solution { public double myPow(double x, int n) { if(n==0) return 1; double mid = myPow(x,n/2); if(n%2==0) return mid*mid; else{ if(n>0) return mid*mid*x; else return mid*mid*(1/x); } } } 分治法,时间复杂度O(nlogn)